Understanding a proof: Artinian ring implies semilocal
Let $R$ be an artinian ring. Then
$R$ is semilocal.
$(operatorname{rad}(R))^n=0$ for some $n$.
Proof: Let $I=text{rad}(R)$. By hyperref[3.15]{3.15}, $exists n$ so that we have $operatorname{ann}_{overline{R}} overline{I}=overline{0}$, where we denote
$overline{I}subset overline{R} = R / (0:I^n)$. We claim $overline{R}=0$. ($Rightarrow 0:I^n=R Rightarrow I^n=0$).
Suppose $overline{R}neq 0$, then since $overline{R}$ is artinian, $exists N$ minimal with the property $Nneq 0$. Now $N$ is simple. Thus $N$ is finite, and $overline{I}N=0$ or $overline{I}N=N.$ The first equation is ruled out by hyperref[3.15]{3.15}, the second one by hyperref[2.2]{2.2} Nakayama's Lemma, ($I=text{rad}(R)$), $N$ finite $R$-module.
I am going over my notes from last semester and I am trying to make the distinction of the proof of 1st bullet point and the 2nd (if there is one). Can someone help me with that? I am providing below the results that I am referring to.
3.15: Let $R$ be a ring that is artinian or noetherian. $I$ an $R$-ideal. Then $0:I^n=0:I^{n+1}$ for some $n$. For such $n$, write
$ overline{R} = left. R middle/ (0:I^n) right.$ and $overline{I}=IR$.
Then $operatorname{ann}_{overline{R}} hspace{0.1cm} overline{I}=0$.
2.2: (Nakayama's Lemma) $M$ a finite $R$-module, $I$ an $R$-ideal. If $M=IM$, then $aM=0$ for some $ain 1+I$. In particular, if $M=IM$ and $Isubset operatorname{Rad}(I)$, then $M=0$ (since $ain R$).
commutative-algebra
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Username UnknownUsername Unknown
1,24742055
add a comment |
Let $R$ be an artinian ring. Then
$R$ is semilocal.
$(operatorname{rad}(R))^n=0$ for some $n$.
Proof: Let $I=text{rad}(R)$. By hyperref[3.15]{3.15}, $exists n$ so that we have $operatorname{ann}_{overline{R}} overline{I}=overline{0}$, where we denote
$overline{I}subset overline{R} = R / (0:I^n)$. We claim $overline{R}=0$. ($Rightarrow 0:I^n=R Rightarrow I^n=0$).
Suppose $overline{R}neq 0$, then since $overline{R}$ is artinian, $exists N$ minimal with the property $Nneq 0$. Now $N$ is simple. Thus $N$ is finite, and $overline{I}N=0$ or $overline{I}N=N.$ The first equation is ruled out by hyperref[3.15]{3.15}, the second one by hyperref[2.2]{2.2} Nakayama's Lemma, ($I=text{rad}(R)$), $N$ finite $R$-module.
I am going over my notes from last semester and I am trying to make the distinction of the proof of 1st bullet point and the 2nd (if there is one). Can someone help me with that? I am providing below the results that I am referring to.
3.15: Let $R$ be a ring that is artinian or noetherian. $I$ an $R$-ideal. Then $0:I^n=0:I^{n+1}$ for some $n$. For such $n$, write
$ overline{R} = left. R middle/ (0:I^n) right.$ and $overline{I}=IR$.
Then $operatorname{ann}_{overline{R}} hspace{0.1cm} overline{I}=0$.
2.2: (Nakayama's Lemma) $M$ a finite $R$-module, $I$ an $R$-ideal. If $M=IM$, then $aM=0$ for some $ain 1+I$. In particular, if $M=IM$ and $Isubset operatorname{Rad}(I)$, then $M=0$ (since $ain R$).
commutative-algebra
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Username UnknownUsername Unknown
1,24742055
add a comment |
Let $R$ be an artinian ring. Then
$R$ is semilocal.
$(operatorname{rad}(R))^n=0$ for some $n$.
Proof: Let $I=text{rad}(R)$. By hyperref[3.15]{3.15}, $exists n$ so that we have $operatorname{ann}_{overline{R}} overline{I}=overline{0}$, where we denote
$overline{I}subset overline{R} = R / (0:I^n)$. We claim $overline{R}=0$. ($Rightarrow 0:I^n=R Rightarrow I^n=0$).
Suppose $overline{R}neq 0$, then since $overline{R}$ is artinian, $exists N$ minimal with the property $Nneq 0$. Now $N$ is simple. Thus $N$ is finite, and $overline{I}N=0$ or $overline{I}N=N.$ The first equation is ruled out by hyperref[3.15]{3.15}, the second one by hyperref[2.2]{2.2} Nakayama's Lemma, ($I=text{rad}(R)$), $N$ finite $R$-module.
I am going over my notes from last semester and I am trying to make the distinction of the proof of 1st bullet point and the 2nd (if there is one). Can someone help me with that? I am providing below the results that I am referring to.
3.15: Let $R$ be a ring that is artinian or noetherian. $I$ an $R$-ideal. Then $0:I^n=0:I^{n+1}$ for some $n$. For such $n$, write
$ overline{R} = left. R middle/ (0:I^n) right.$ and $overline{I}=IR$.
Then $operatorname{ann}_{overline{R}} hspace{0.1cm} overline{I}=0$.
2.2: (Nakayama's Lemma) $M$ a finite $R$-module, $I$ an $R$-ideal. If $M=IM$, then $aM=0$ for some $ain 1+I$. In particular, if $M=IM$ and $Isubset operatorname{Rad}(I)$, then $M=0$ (since $ain R$).
commutative-algebra
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Username UnknownUsername Unknown
1,24742055
Let $R$ be an artinian ring. Then
$R$ is semilocal.
$(operatorname{rad}(R))^n=0$ for some $n$.
Proof: Let $I=text{rad}(R)$. By hyperref[3.15]{3.15}, $exists n$ so that we have $operatorname{ann}_{overline{R}} overline{I}=overline{0}$, where we denote
$overline{I}subset overline{R} = R / (0:I^n)$. We claim $overline{R}=0$. ($Rightarrow 0:I^n=R Rightarrow I^n=0$).
Suppose $overline{R}neq 0$, then since $overline{R}$ is artinian, $exists N$ minimal with the property $Nneq 0$. Now $N$ is simple. Thus $N$ is finite, and $overline{I}N=0$ or $overline{I}N=N.$ The first equation is ruled out by hyperref[3.15]{3.15}, the second one by hyperref[2.2]{2.2} Nakayama's Lemma, ($I=text{rad}(R)$), $N$ finite $R$-module.
I am going over my notes from last semester and I am trying to make the distinction of the proof of 1st bullet point and the 2nd (if there is one). Can someone help me with that? I am providing below the results that I am referring to.
3.15: Let $R$ be a ring that is artinian or noetherian. $I$ an $R$-ideal. Then $0:I^n=0:I^{n+1}$ for some $n$. For such $n$, write
$ overline{R} = left. R middle/ (0:I^n) right.$ and $overline{I}=IR$.
Then $operatorname{ann}_{overline{R}} hspace{0.1cm} overline{I}=0$.
2.2: (Nakayama's Lemma) $M$ a finite $R$-module, $I$ an $R$-ideal. If $M=IM$, then $aM=0$ for some $ain 1+I$. In particular, if $M=IM$ and $Isubset operatorname{Rad}(I)$, then $M=0$ (since $ain R$).
commutative-algebra
commutative-algebra
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Username UnknownUsername Unknown
1,24742055
edited 2 days ago
Bernard
118k639112
asked 2 days ago
Username UnknownUsername Unknown
1,24742055
edited 2 days ago
Bernard
118k639112
edited 2 days ago
Bernard
118k639112
edited 2 days ago
Bernard
118k639112
118k639112
asked 2 days ago
Username UnknownUsername Unknown
1,24742055
asked 2 days ago
Username UnknownUsername Unknown
1,24742055
asked 2 days ago
Username UnknownUsername Unknown
1,24742055
1,24742055
add a comment |
add a comment |
1 Answer
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Everything you have written seems to be geared to proving the second point. I see nothing about the first point.
The first point is trivial though, so maybe you didn’t write it down? $R/J(R)$ has Jacobson radical zero, and is artinian since $R$ is. Therefore it’s semisimple, so $R$ is semilocal.
answered 2 days ago
rschwiebrschwieb
105k12100245
This can also be proved from "first principles" i.e. without knowing about semisimple rings. Such a proof is in Atiyah-Macdonald.
– RghtHndSd
2 days ago
@RghtHndSd Well, if your definition of semilocal is $R/J(R)$ is semisimple” it might be hard to do without semisimple rings , but I suppose in this case his definition might be “finitely many maximal ideals?” Those are the two I’m aware of. Is that what’s in A-M?
– rschwieb
2 days ago
Yes - it's Proposition 8.3 in A-M. Consider the collection of all finite intersections of maximal ideals in $R$. $R$ artin implies that this collection has a minimal element, say $I = m_1 cap dots cap m_k$. For any maximal ideal $m$, $m cap I = I$ by minimality, and so $I subset m$. By a straightforward element argument (A-M 1.11b), this implies $m_i subset m$ for some $i$. That $m_i$ is maximal finishes the argument.
– RghtHndSd
yesterday
add a comment |
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Everything you have written seems to be geared to proving the second point. I see nothing about the first point.
The first point is trivial though, so maybe you didn’t write it down? $R/J(R)$ has Jacobson radical zero, and is artinian since $R$ is. Therefore it’s semisimple, so $R$ is semilocal.
answered 2 days ago
rschwiebrschwieb
105k12100245
This can also be proved from "first principles" i.e. without knowing about semisimple rings. Such a proof is in Atiyah-Macdonald.
– RghtHndSd
2 days ago
@RghtHndSd Well, if your definition of semilocal is $R/J(R)$ is semisimple” it might be hard to do without semisimple rings , but I suppose in this case his definition might be “finitely many maximal ideals?” Those are the two I’m aware of. Is that what’s in A-M?
– rschwieb
2 days ago
Yes - it's Proposition 8.3 in A-M. Consider the collection of all finite intersections of maximal ideals in $R$. $R$ artin implies that this collection has a minimal element, say $I = m_1 cap dots cap m_k$. For any maximal ideal $m$, $m cap I = I$ by minimality, and so $I subset m$. By a straightforward element argument (A-M 1.11b), this implies $m_i subset m$ for some $i$. That $m_i$ is maximal finishes the argument.
– RghtHndSd
yesterday
add a comment |
Everything you have written seems to be geared to proving the second point. I see nothing about the first point.
The first point is trivial though, so maybe you didn’t write it down? $R/J(R)$ has Jacobson radical zero, and is artinian since $R$ is. Therefore it’s semisimple, so $R$ is semilocal.
answered 2 days ago
rschwiebrschwieb
105k12100245
This can also be proved from "first principles" i.e. without knowing about semisimple rings. Such a proof is in Atiyah-Macdonald.
– RghtHndSd
2 days ago
@RghtHndSd Well, if your definition of semilocal is $R/J(R)$ is semisimple” it might be hard to do without semisimple rings , but I suppose in this case his definition might be “finitely many maximal ideals?” Those are the two I’m aware of. Is that what’s in A-M?
– rschwieb
2 days ago
Yes - it's Proposition 8.3 in A-M. Consider the collection of all finite intersections of maximal ideals in $R$. $R$ artin implies that this collection has a minimal element, say $I = m_1 cap dots cap m_k$. For any maximal ideal $m$, $m cap I = I$ by minimality, and so $I subset m$. By a straightforward element argument (A-M 1.11b), this implies $m_i subset m$ for some $i$. That $m_i$ is maximal finishes the argument.
– RghtHndSd
yesterday
add a comment |
Everything you have written seems to be geared to proving the second point. I see nothing about the first point.
The first point is trivial though, so maybe you didn’t write it down? $R/J(R)$ has Jacobson radical zero, and is artinian since $R$ is. Therefore it’s semisimple, so $R$ is semilocal.
answered 2 days ago
rschwiebrschwieb
105k12100245
Everything you have written seems to be geared to proving the second point. I see nothing about the first point.
The first point is trivial though, so maybe you didn’t write it down? $R/J(R)$ has Jacobson radical zero, and is artinian since $R$ is. Therefore it’s semisimple, so $R$ is semilocal.
answered 2 days ago
rschwiebrschwieb
105k12100245
answered 2 days ago
rschwiebrschwieb
105k12100245
answered 2 days ago
rschwiebrschwieb
105k12100245
answered 2 days ago
rschwiebrschwieb
105k12100245
105k12100245
This can also be proved from "first principles" i.e. without knowing about semisimple rings. Such a proof is in Atiyah-Macdonald.
– RghtHndSd
2 days ago
@RghtHndSd Well, if your definition of semilocal is $R/J(R)$ is semisimple” it might be hard to do without semisimple rings , but I suppose in this case his definition might be “finitely many maximal ideals?” Those are the two I’m aware of. Is that what’s in A-M?
– rschwieb
2 days ago
Yes - it's Proposition 8.3 in A-M. Consider the collection of all finite intersections of maximal ideals in $R$. $R$ artin implies that this collection has a minimal element, say $I = m_1 cap dots cap m_k$. For any maximal ideal $m$, $m cap I = I$ by minimality, and so $I subset m$. By a straightforward element argument (A-M 1.11b), this implies $m_i subset m$ for some $i$. That $m_i$ is maximal finishes the argument.
– RghtHndSd
yesterday
add a comment |
This can also be proved from "first principles" i.e. without knowing about semisimple rings. Such a proof is in Atiyah-Macdonald.
– RghtHndSd
2 days ago
@RghtHndSd Well, if your definition of semilocal is $R/J(R)$ is semisimple” it might be hard to do without semisimple rings , but I suppose in this case his definition might be “finitely many maximal ideals?” Those are the two I’m aware of. Is that what’s in A-M?
– rschwieb
2 days ago
Yes - it's Proposition 8.3 in A-M. Consider the collection of all finite intersections of maximal ideals in $R$. $R$ artin implies that this collection has a minimal element, say $I = m_1 cap dots cap m_k$. For any maximal ideal $m$, $m cap I = I$ by minimality, and so $I subset m$. By a straightforward element argument (A-M 1.11b), this implies $m_i subset m$ for some $i$. That $m_i$ is maximal finishes the argument.
– RghtHndSd
yesterday
This can also be proved from "first principles" i.e. without knowing about semisimple rings. Such a proof is in Atiyah-Macdonald.
– RghtHndSd
2 days ago
This can also be proved from "first principles" i.e. without knowing about semisimple rings. Such a proof is in Atiyah-Macdonald.
– RghtHndSd
2 days ago
@RghtHndSd Well, if your definition of semilocal is $R/J(R)$ is semisimple” it might be hard to do without semisimple rings , but I suppose in this case his definition might be “finitely many maximal ideals?” Those are the two I’m aware of. Is that what’s in A-M?
– rschwieb
2 days ago
@RghtHndSd Well, if your definition of semilocal is $R/J(R)$ is semisimple” it might be hard to do without semisimple rings , but I suppose in this case his definition might be “finitely many maximal ideals?” Those are the two I’m aware of. Is that what’s in A-M?
– rschwieb
2 days ago
Yes - it's Proposition 8.3 in A-M. Consider the collection of all finite intersections of maximal ideals in $R$. $R$ artin implies that this collection has a minimal element, say $I = m_1 cap dots cap m_k$. For any maximal ideal $m$, $m cap I = I$ by minimality, and so $I subset m$. By a straightforward element argument (A-M 1.11b), this implies $m_i subset m$ for some $i$. That $m_i$ is maximal finishes the argument.
– RghtHndSd
yesterday
Yes - it's Proposition 8.3 in A-M. Consider the collection of all finite intersections of maximal ideals in $R$. $R$ artin implies that this collection has a minimal element, say $I = m_1 cap dots cap m_k$. For any maximal ideal $m$, $m cap I = I$ by minimality, and so $I subset m$. By a straightforward element argument (A-M 1.11b), this implies $m_i subset m$ for some $i$. That $m_i$ is maximal finishes the argument.
– RghtHndSd
yesterday
add a comment |
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